Solutions to Math 381 Quiz 2

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1 Solutions to Math 381 Quiz 2 November 7, 2018 (1) In one sentence, what is the goal of your class project? Each team has a different answer here. (2) Write a function in Python which computes a linear function of 3 numerical inputs and returns 170 on the input 2, 3, 7. def foo(a,b,c): return 80*a+b+c; foo(2,3,7) 170 (3) Amazon s CEO Jeff Bezos is touring cities in America looking for the best place to build the new HQ2 (second headquarters). There are currently 20 cities on the list. Bezos wants you to find the shortest tour visiting all 20 cities, starting and ending in Seattle. Assume all distances between pairs of cities are given by a matrix D = (d ij ). To get started on the job, how would you translate this problem into an integer programming problem in standard form? Recall that standard form of an LP requires that it (a) Must be a maximization problem. (b) All constraints in the form i a ix i b. (c) x i 0 for all i Since we are doing integer programming, we also require our variables be integers. Notice that there are actually 21 cities in this problem! Thus D is There are a few different ways to model this problem. I will outline how they work. Method 1. Consider first modeling the problem as a directed graph. In this case, we can define (for 1 i, j 21) { 1, The path travels from city i to city j, x ij = 0, otherwise to be our decision variables (note that they satisfy condition (c) above). Then we want to maximize C = d ij x ij i=1 1

2 2 subject to the constraints i=1 i=1 S [21], x ij S 1 i,j S The last constraint (actually this is constraints in compact notation!) was the most difficult for most people and caused them the most grief. This is what I called the subtour constraint and it is what makes it so that the eventual path is connected. For instance, notice that the directed graph on six vertices: B A C D E F satisfies the first four constraints on in- and out-degree but clearly is not a solution. One last thing to mention here is that many of you added the first four constraints as well as some equations that were equivalent to x ij = 21 i j but if you look carefully, that is actually already implied by the first four. Method 2. The second method is similar but uses a slight trick which reduces the number of variables we have to consider. This time we model the problem as an undirected graph. Our variables are defined the same way but this time we consider x ij = x ji. The idea here is that it really doesn t matter what way we are traversing the tour, just that we have a tour at all. Since x ii = 0, notice that really we are only working with the entries in the upper triangular matrix (x ij ). This gives us (n 1) = n(n 1) 2

3 3 variables to optimize instead of n 2 (fewer than half). The rest is more-or-less the same, except now our objective function is on fewer variables: i=1 j=i+1 d ij x ij and since we have no notion of in- and out-degree, the constraints become x ij 2, x ij 2 S [21], x ij S 1 i,j S Which is still quite a few constraints to encode. Method 3. Another tradeoff you can make (instead of reducing the number of variables) is to reduce the number of constraints by absorbing some of the complexity into your variables and objective function. At the end of the day there is no way to solve TSP in a small, simple system because if you could you d be the proud recipient of a million dollars for solving P = NP. So maybe this is the right way to go. The trick here is to define your variables differently: for 1 i, j 21, { 1, The path visits city i at step j, x ij = 0, otherwise where we say here that step 1 is the city visited directly after Seattle the first time. Now if we say that city 1 is Seattle, we require that x 1 21 = 1 and x 1,j = 0 otherwise, so we can actually remove the variables x 1,i and x i,21 from the problem completely. Then our objective function becomes k=1 x i k x j (k+1) d ij i=2 j=2 since we are interested in counting the distance d ij if and only if there is a k such that we visit i on the k th step and j on the (k + 1) st. The upshot here is that by the nature of the variables, we can ensure there are no proper cycles as long as whenever x ij = x kl = 1 and j l, then i k. In

4 4 other words, we never visit the same city twice. Our constraints, then, are i=2 i= i=1 Where now these represent 400 constraints. This might seem like a lot, but remember that n = 20 here so the number of constraints is O(n 2 ) where before its was O(2 n ). (4) The Faster-Than-Fast Internet company has asked Math 381 Inc to identify clusters of cities to connect using their new long fiber optic cables. These cables must start and end at the main hub in one of the cities. They can only lay the cables along the routes given in a network map such as the one we saw including San Francisco and Salt Lake City. The cables cannot be longer than 150 miles, or they would be unreliable. The company would like to minimize the number of clusters in their network because each one requires a separate electrical generator. In each connected cluster, they also want to minimize the total length of cable used. (a) What is the name of the algorithm you suggest they use? (b) Give a general description of the algorithm. (c) Justify your suggestion. (a) If the goal is to connect the cities in such a way as to minimize the total length of cable used, then either Kruskal s algorithm or Prim s algorithm would be helpful if the cities are not too far apart so that there exists a spanning tree on the complete graph with edge labels given by distance and all cites are connected via edges with weight at most 150. Let s build on one of those two algorithms. I chose Kruskal s algorithm. (b) To run the algorithm, assume we are given the weighted adjacency matrix W = (w i,j ) for the complete graph on all cities under consideration. If we can go directly from city i to city j, then w i,j is the distance along the route, if not then we assume w i,j is set to any number bigger than 150, note 151 will suffice. Cluster the cities by declaring that city i and city j are in the same cluster if w i,j 150, so every pair of cities in one cluster is connected via an allowed route using only cables of length at most 150 miles. Note, any two cities in different clusters are more than 150 miles apart. Now, on each cluster, we can run Kruskal s algorithm on each cluster. Recall, Kruscal s algorithm starts by sorting the edges connecting any two cities in

5 5 the same cluster by weight, adding the smallest edge first. Continue adding edges in increasing order of weight from the same cluster provided they do not create a cycle. Once all of the edges up to weight 150 are included, we must have connected every vertex in given cluster. Thus, we identify a minimal weight spanning tree on each cluster. (c) By clustering cities as above, we have minized the total number of clusters by making every cluster as big as possible given the cable length constraint. By choosing a minimal weight spanning tree on each cluster we have minimized the total length of cable used. Thus, this algorithm will achieve an optimal solution for where to lay the cable. (5) The Stable Match Algorithm was described in class for n students and n hospitals. In reality, the number of students is not the same as the number of hospitals. How would you modify the algorithm to work for s students and h hospitals for any positive integers s and h so that there are as many matches as possible and no unstable pairs. As most of you noticed there is very little you have to do to make the Stable Match Algorithm work for uneven sets of students. One tactic which simplifies the problem is to notice that we never asked that one group always propose to the other. Therefore you can assume without loss of generality that s < h and create an algorithm that works. Then if h < s you can just flip everything. You can do one of two things: first, you can try just running the algorithm as you normally would with students making the offers. Since all the students rank all the hospitals and since h > s, eventually every student will make an offer to a hospital that has no other offers (every round at least one new offer is made, so on the s th iteration there must be s provisional acceptances, but this means that every student must be in one so we are done) and at this point we terminate the process. This proves that there is some algorithm which finds some matching but to get full credit here I expected you to say something about why the solution it finds is still maximal and stable. The maximality is relatively clear in the end state every student has been matched with a hospital and that is the most matching we could hope for when s < h. The stability can be seen by the following: consider any hospital H j that was not matched in the matching M above. The only way this could happen is if no student ever proposes to H j, thus no student prefers H j to the hospital they have been assigned in M. Thus the unmatched hospitals can never be part of an unstable pair in M. Therefore if there were an unstable pair, it would have to be among the matches that were made. Say S i is matched to H j in M, but S j prefers H k over H j. Then during the algorithm, S i proposed to H k prior to proposing to H j and was rejected by H k in favor of another student. Since hospitals only move up in preference as the algorithm progresses, we know that H k prefers their assignment in M over S j. Hence, S j is not part of an unstable pair, for any student S j. The other method you can use to a similar end is padding the students out with s h dummy students. You must require that every dummy student is

6 preferred below the other students by every hospital. Then after running the (unaltered) algorithm you can take your matching and throw out any matching that includes a dummy student.

6 6 preferred below the other students by every hospital. Then after running the (unaltered) algorithm you can take your matching and throw out any matching that includes a dummy student. By construction if a hospital was matched with a dummy student, no real student can prefer that hospital more than the one they are already matched with. Thus no instabilities can arise from the unmatched hospitals. The stability of the matched pairs follows since every subset of a stable matching is stable. Again the maximality follows since every student ends up matched to a hospital. To learn more about this amazing algorithm, read the original paper by Gale and Shapley from 1962 called College Admissions and the Stability of Marriage. It s a gem! (6) This page is intentionally left blank. If you finish the quiz early, you could either work on your project here or draw a picture of Discrete Math Modeling in Action to be considered as the new cover of the course notes. No points will be added or subtracted based on this page.

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